MA 3416: Group representations
Selected answers/solutions to the assignment due March 23, 2015
1. (a) Indeed, scalar matrices do not change under conjugation by anything, in particular
by matrices of representation, so if g is in Gρ , then hgh−1 is in Gρ as well.
(b) Since trace of an operator is equal to the sum of its eigenvalues, and all eigenvalues
are roots of unity, our claim is clear.
(c) For complex numbers λ1 , . . . , λn we have |λ1 + . . . + λn | = |λ1 | + . . . + |λn | if and only
if the vectors representing these numbers in 2d plane have the same direction. Given that all
these numbers are roots of unity in our case, we conclude that they are all equal, and so are
eigenvalues of a scalar matrix.
2. We shall use both intertwining numbers and characters to show how they all work.
(a) Clearly, this representation is isomorphic to the standard representation on C4 , and
hence is isomorphic to the sum of the trivial representation and V, the constant functions
and functions that add up to zero.
(b) The intertwining number of this representation with itself is equal to three, since there
are three different orbits on pairs of edges: two edges can coincide, have a common vertex,
or be disjoint. Also, the intertwining number of this representation with the representation
arising from the set of vertices is two: in a pair of a vertex and an edge, the vertex can
belong to the edge or not. Therefore, this representation contains a copy of the trivial one,
a copy of V, and something else. That something else can only be U, as it is the only twodimensional representation. The three invariant subspaces where those representations are
realised are constant functions, functions that take opposite values on pairs of disjoint edges,
and functions that take the same values on pairs of disjoint edges and add up to zero.
(c) Clearly, this representation is isomorphic to the standard representation on C4 , and
hence is isomorphic to the sum of the trivial representation and V, the constant functions
and functions that add up to zero.
(d) Let us augment the character table of S4 with the values of character of the corresponding representations:
triv
sign
V
V0
U
edges
e
1
1
3
3
2
12
(ij)
1
-1
1
-1
0
2
(ijk)
1
1
0
0
-1
0
(ijkl)
1
-1
-1
1
0
0
(ij)(kl)
1
1
-1
-1
2
0
Computing the scalar products with irreducible characters, we see that this representation is
isomorphic to triv ⊕ V ⊕ V ⊕ V 0 ⊕ U.
Let us note that the space of all even functions on edges has three natural invariant
subspaces: constant functions, functions that are constant on each of the three directions and
add up to zero, and functions that take opposite values on each two edges of the same face
that are not connected with an edge. It remains to find two invariant subspaces in the space
of odd functions to guarantee that we found all the irreducible constituents. Those invariant
subspaces consist of functions for which the sum of values on edges in each face is zero and of
functions for which for each face the sum of values of two opposite edges is equal to the sum
of edges on two other opposite edges.
3. In class, we proved that A5 is the group of all rotations of the dodecahedron, and
identified the conjugacy classes. Since the only rotations of the dodecahedron preserving
some face are rotations about the axis passing through the centre of that face, only cycles of
length 5 have fixed points in the set of faces, and we get the following character of CM:
#
CM
15
1
12
22 1
15
0
312
20
0
51
12
2
52
12
2
Using the character table
#
triv
V
W
V1
V2
from class (here τ =
√
1+ 5
2 ),
15
1
1
4
5
3
3
22 1
15
1
0
1
-1
-1
312
20
1
1
-1
0
0
51
12
1
-1
0
τ
1−τ
52
12
1
-1
0
1−τ
τ
and computing scalar products, we see that
CM ' triv ⊕ V1 ⊕ V2 ⊕ W.
4. It is clear that W1 ⊗ W2 ' W2 ⊗ W1 and W ⊗ triv ' W, so we just need to decompose
10 pairwise products whose characters are
#
V1 ⊗ V1
V1 ⊗ V2
V1 ⊗ V
V1 ⊗ W
V2 ⊗ V2
V2 ⊗ V
V2 ⊗ W
V ⊗V
V ⊗W
W⊗W
15
1
9
9
12
15
9
12
15
16
20
25
22 1
15
1
1
0
-1
1
-1
-1
0
0
1
312
20
0
0
0
0
0
0
0
1
-1
1
51
12
τ+1
-1
−τ
0
2−τ
−1 + τ
0
1
0
0
52
12
2−τ
-1
−1 + τ
0
τ+1
−τ
0
1
0
0
Computing scalar products of characters, we obtain
V1 ⊗ V1 ' triv ⊕ V1 ⊕ W
V1 ⊗ V2 ' V ⊕ W
V1 ⊗ V ' V2 ⊕ V ⊕ W
V1 ⊗ W ' V1 ⊕ V2 ⊕ V ⊕ W
V2 ⊗ V2 ' triv ⊕ V2 ⊕ W
V2 ⊗ V ' V1 ⊕ V ⊕ W
V2 ⊗ W ' V1 ⊕ V2 ⊕ V ⊕ W
V ⊗ V ' triv ⊕ V1 ⊕ V2 ⊕ V ⊕ W
V ⊗ W ' V1 ⊕ V2 ⊕ V ⊕ W ⊕ W
W ⊗ W ' triv ⊕ V1 ⊕ V2 ⊕ V ⊕ V ⊕ W ⊕ W
Note that since under conjugation by a transposition V1 and V2 get swapped, and other
representations remain unchanged, V1 ⊗ V ' V2 ⊕ V ⊕ W implies V2 ⊗ V ' V1 ⊕ V ⊕ W etc.
5. (a) Clearly (g1 , g2 )(h1 , h2 )(g1 , g2 )−1 ) = (g1 h1 g1 −1, g2 h2 g−1
2 ), from which it is obvious
that conjugacy classes of G1 × G2 are C1 × C2 , where Ci is a conjugacy class of Gi .
(b) Let us compute the scalar product of such a character with itself:
(χ, χ) =
X
1
|χ((g1 , g2 ))|2 =
#G1 #G2
(g1 ,g2 )
=
X
1
| tr(ρ1 (g1 ) ⊗ ρ2 (g2 ))|2 =
#G1 #G2
(g1 ,g2 )
X
1
=
| tr(ρ1 (g1 ))|2 | tr(ρ2 (g2 ))|2 =
#G1 #G2
(g1 ,g2 )
1 X
1 X
=
| tr(ρ1 (g1 ))|2
| tr(ρ2 (g2 ))|2 = (χ1 , χ1 )(χ2 , χ2 ) = 1,
#G1 g
#G2 g
1
2
so χ is an irreducible character. A similar computation shows that these characters are
pairwise orthogonal. Also, their number is equal to the product of numbers of conjugacy
classes in our groups which by (a) is the number of conjugacy classes in the product. Thus,
these are all irreducible representations.